Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.1.68b

Chapter 7, Problem 7.1.68b

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

68. y= (3x+2)/(2x-11), -2 ≤ x ≤ 2, x_0=1/2

Verified step by step guidance

Start with the given function: \(y = \frac{3x + 2}{2x - 11}\), where \(-2 \leq x \leq 2\).

To find the inverse function \(g(y)\), first solve the equation \(y = \frac{3x + 2}{2x - 11}\) for \(x\) in terms of \(y\).

Multiply both sides of the equation by the denominator \((2x - 11)\) to eliminate the fraction: \(y(2x - 11) = 3x + 2\).

Distribute \(y\) on the left side: \(2xy - 11y = 3x + 2\).

Group all terms involving \(x\) on one side and constants on the other: \(2xy - 3x = 11y + 2\). Then factor out \(x\): \(x(2y - 3) = 11y + 2\). Finally, solve for \(x\): \(x = \frac{11y + 2}{2y - 3}\). This expression defines the inverse function \(g(y)\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Functions

An inverse function reverses the effect of the original function, swapping inputs and outputs. For a function f(x), its inverse g(y) satisfies g(f(x)) = x. Finding the inverse involves solving y = f(x) for x in terms of y, which is essential for understanding how to express x as a function of y.

Domain and Range Restrictions

When finding inverses, it is crucial to consider the domain and range of the original function to ensure the inverse is well-defined and one-to-one. Restricting the domain, as in -2 ≤ x ≤ 2, helps avoid ambiguity and guarantees the inverse function exists and behaves properly over that interval.

Derivative and Tangent Line Approximation

The derivative of a function at a point gives the slope of the tangent line there, which can be used for linear approximations. Understanding how to compute derivatives of both the function and its inverse is key to analyzing behavior near specified points, such as x₀ = 1/2, and for approximating values.

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Slopes of Tangent Lines

Related Practice

Textbook Question

Evaluate the integrals in Exercises 67–74 in terms of

b. natural logarithms.

71. ∫(from 1/5 to 3/13)dx/(x√(1-16x²))

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Textbook Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

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Textbook Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.

4. b. arcsin(-1/√2)

Textbook Question

75. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.

g(x) = x(ln x)²

Textbook Question

Find the volumes of the solids in Exercises 135 and 136.

135. The solid lies between planes perpendicular to the x-axis at x=-1 and x=1. The cross-sections perpendicular to the x-axis are

b. vertical squares whose base edges run from the curve y=-1/√(1+x²) to the curve y=1/√(1+x²).

Textbook Question

3. Use the properties of logarithms to write the expressions in Exercises 3 and 4 as a single term.

b. ln(3x² - 9x) + ln(1/3x)